Having problem solving a limit

[the_morbidus]

having problem solving a limit!

Homework Statement

Solve the following limits algebraically.

lim x->5 x^2 - 25 / √(2x+6) -4

Homework Equations

The Attempt at a Solution

i've tried a different few ways.

so by factoring the top i have (x+5)(x-5) so i get x=5 or x=-5, and while 5 is the limit, -5 on top with the square it will just render the negative sign useless.

tried to multiply the bottom cognitive
(x^2-25)(√(2x+6)+4)/(√(2x+6)-4)(√(2x+6)+4) =
=(x^2-25)(√(2x+6)+4)/(√(2x+6))^2 - (4)^2 =
=(x^2-25)(√(2x+6)+4)/2x+6-16=
=(x^2-25)(√(2x+6)+4)/2x-10

i'm stuck here and i doubt this is even the proper route.

 

[Mark44]

Homework Statement

Solve the following limits algebraically.

lim x->5 x^2 - 25 / √(2x+6) -4

Homework Equations

The Attempt at a Solution

i've tried a different few ways.

so by factoring the top i have (x+5)(x-5) so i get x=5 or x=-5, and while 5 is the limit, -5 on top with the square it will just render the negative sign useless.

tried to multiply the bottom cognitive
(x^2-25)(√(2x+6)+4)/(√(2x+6)-4)(√(2x+6)+4) =
=(x^2-25)(√(2x+6)+4)/(√(2x+6))^2 - (4)^2 =
=(x^2-25)(√(2x+6)+4)/2x+6-16=
=(x^2-25)(√(2x+6)+4)/2x-10

i'm stuck here and i doubt this is even the proper route.

Multiplying by the conjugate over itself is a useful approach.

[tex]\lim_{x \to 5}\frac{(x-5)(x+5)}{\sqrt{2x+6}-4}\cdot \frac{\sqrt{2x+6} + 4}{\sqrt{2x+6} + 4}[/tex]
[tex]= \lim_{x \to 5}\frac{(x-5)(x+5)(\sqrt{2x+6} + 4)}{(\sqrt{2x+6}-4) \cdot (\sqrt{2x+6} + 4)}[/tex]
Can you take it from here?

 

[symbolipoint]

Solve the following limits algebraically.

lim x->5 x^2 - 25 / √(2x+6) -4

What you really mean is
[tex]Lim_{x\rightarrow 5} \frac{x^2-25}{\sqrt{2x+6}-4}[/tex]

...

Most of your work looks good. Try then dividing numerator and denominator by either x, or x^2 ?

 

[symbolipoint]

Multiplying by the conjugate over itself is a useful approach.

[tex]\lim_{x \to 5}\frac{(x-5)(x+5)}{\sqrt{2x+6}-4}\cdot \frac{\sqrt{2x+6} + 4}{\sqrt{2x+6} + 4}[/tex]
[tex]= \lim_{x \to 5}\frac{(x-5)(x+5)(\sqrt{2x+6} + 4)}{(\sqrt{2x+6}-4)\cdot (\sqrt{2x+6} + 4)}[/tex]
Can you take it from here?

He appeared to be doing that; maybe I was confused looking at all the steps and symbols through text?
EDIT: Give me the chance to try the problem myself; maybe more comment later.

OK, very neat. Just multiplying numerator and denominator by conjugate gives you something to simplify and then simple substitution of x=5 can be evaluated with no complications.

 

[HallsofIvy]

Homework Statement

Solve the following limits algebraically.

lim x->5 x^2 - 25 / √(2x+6) -4

Homework Equations

The Attempt at a Solution

i've tried a different few ways.

so by factoring the top i have (x+5)(x-5) so i get x=5 or x=-5, and while 5 is the limit, -5 on top with the square it will just render the negative sign useless.

tried to multiply the bottom cognitive
(x^2-25)(√(2x+6)+4)/(√(2x+6)-4)(√(2x+6)+4) =
=(x^2-25)(√(2x+6)+4)/(√(2x+6))^2 - (4)^2 =
=(x^2-25)(√(2x+6)+4)/2x+6-16=
=(x^2-25)(√(2x+6)+4)/2x-10

i'm stuck here and i doubt this is even the proper route.

Both numerator and denominator are 0 when x= 5 therefore both have a factor of x- 5:
[tex]\frac{(x- 5)(x+ 5)(\sqrt{2x+ 6}+ 4)}{2(x- 5)}[/tex]

 

[the_morbidus]

thank you guys, when I woke up this morning and having looked at a few other problems I arrived at the same solution as HallsofIvy, I then removed the (x-5) and substituted the x for 5 and i got 40 as the limit I think.

 

[symbolipoint]

thank you guys, when I woke up this morning and having looked at a few other problems I arrived at the same solution as HallsofIvy, I then removed the (x-5) and substituted the x for 5 and i got 40 as the limit I think.

Yes, and in fact, you both think and KNOW for sure now.